blocks|key|87580|text|def+fib(n):
++++tup=[]
++++a,b+=+0,1
++++while+b<n:
++++++++tup=tup%2B[b,]
++++++++a,b+=+b,a%2Bb
++++print+tup|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|87581|您需要附加到元组，然后如果您愿意的话打印它。|unstyled|87582|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|J|8|@]|9|@]|A|$]]|$1|G|3|-4|5|F|7|K|8|@]|9|@]|A|$]]]|H|$]]

<pre><code>def fib(n):
 tup=[]
 a,b = 0,1
 while b&lt;n:
 tup=tup+[b,]
 a,b = b,a+b
 print tup
</code></pre>

You need to append to tuple and then print it if you like

blocks|key|3471343|text|您可以将其附加到列表中：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3471344|def+make_fibonacci(n):
++++fib+=+[]
++++a,+b+=+0,+1
++++for+i+in+range(n):
++++++++a,+b+=+b,+a%2Bb
++++++++fib.append(a)
++++return+tuple(fib)


make_fibonacci(10)|code-block|syntax|javascript|3471345|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

You can append it to a list:

<pre><code>def make_fibonacci(n):
 fib = []
 a, b = 0, 1
 for i in range(n):
 a, b = b, a+b
 fib.append(a)
 return tuple(fib)


make_fibonacci(10)
</code></pre>

blocks|key|3471370|text|您可以使用列表来实现这一点：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3471371|def+make_fibonacci(n):
++++result+=+[0]
++++a,+b+=+0,+1
++++for+i+in+range(n-1):
++++++++a,+b+=+b,+a%2Bb
++++++++result.append(b)
++++return+tuple(result)


>>>+print+make_fibonacci(10)
(0,+1,+1,+2,+3,+5,+8,+13,+21,+34,+55)|code-block|syntax|javascript|3471372|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

You can do that using a list:

<pre><code>def make_fibonacci(n):
 result = [0]
 a, b = 0, 1
 for i in range(n-1):
 a, b = b, a+b
 result.append(b)
 return tuple(result)


&gt;&gt;&gt; print make_fibonacci(10)
(0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55)
</code></pre>

blocks|key|3471425|text|下面是一个简单的解决方案，构造为一个列表并返回一个元组|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3471426|def+make_fibonacci(n):
++++a+=+[0,+1]
++++[a.append(a[-1]%2Ba[-2])+for+i+in+xrange(n)]
++++return+tuple(a)|code-block|syntax|javascript|3471427|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

here's a naive solution, construct as a list and return a tuple

<pre><code>def make_fibonacci(n):
 a = [0, 1]
 [a.append(a[-1]+a[-2]) for i in xrange(n)]
 return tuple(a)
</code></pre>

blocks|key|87696|text|这个纯粹的工作在元组上，不受列表限制。|type|unstyled|depth|inlineStyleRanges|offset|length|style|BOLD|entityRanges|data|87697|概念知识：可以添加两个元组|87698|def+get_fibonacci(n):++#+n+=+no.+of+terms+to+be+taken+in+tuple+fibonacci_sequence
++++a,+b+=+0,+1
++++fibonacci_sequence+=+(a,+b)
++++for+i+in+range(n-2):
++++++++a,+b+=+b,+a%2Bb
++++++++fibonacci_sequence+%2B=+(b,)++#+Comma+is+needed+at+the+end+to+accept+a+tuple+with+a+single+value
++++print+fibonacci_sequence
get_fibonacci(10)|code-block|syntax|javascript|87699|Python+：这已经在Python+v2.7.12上进行了测试，它可能无法在Python3.x和更高版本上工作。如果你认为它有效，好的，好的，请纠正这一说法。积极的修改永远欢迎！|87700|entityMap^0|5|5|0|0|4|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@$9|R|A|S|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|T|8|@$9|U|A|V|B|C]]|D|@]|E|$]]|$1|H|3|I|5|J|7|W|8|@]|D|@]|E|$K|L]]|$1|M|3|N|5|6|7|X|8|@]|D|@]|E|$]]|$1|O|3|-4|5|6|7|Y|8|@]|D|@]|E|$]]]|P|$]]

This one purely works on tuples and is free from lists.

Conceptual Knowledge : Two Tuples Can Be Added

<pre><code>def get_fibonacci(n): # n = no. of terms to be taken in tuple fibonacci_sequence
 a, b = 0, 1
 fibonacci_sequence = (a, b)
 for i in range(n-2):
 a, b = b, a+b
 fibonacci_sequence += (b,) # Comma is needed at the end to accept a tuple with a single value
 print fibonacci_sequence
get_fibonacci(10)
</code></pre>

Note: This has been tested on Python Console v2.7.12, and it might not work on Python 3.x and later versions. If you find it working, well and fine, please correct this statement.
 Positive Modifications are always Welcome!

To be writing a make_fibonacci that accepts a parameter n which generates and returns a tuple containing the first n+1 terms of the fibonacci sequence, where n>= 0. From the other questions here, 

<pre><code>def make_fibonacci(n):
 a, b = 0, 1
 for i in range(d):
 a, b = b, a+b
</code></pre>

but since I need the range of the fibonacci in a tuple, like 

<pre><code>make_fibonacci(10) 
&gt;&gt;&gt; (0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55)
</code></pre>

Thanks a lot!

Fibonacci Sequence in tuple Python

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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要编写接受参数n的make_fibonacci，该参数生成并返回包含fibonacci序列的第一个n+1项的元组，其中n>= 0。从其他问题上看，def make_fibonacci(n):    a, b = 0, 1    for i in range(d):        a, b = b, a+b但是既然我需要...

问元组Python中的Fibonacci序列
EN

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问元组Python中的Fibonacci序列EN